Domain of cosine inverse x :
For -1 ≤ x ≤ 1, define cos^{-1} x as the unique number y in [0, π] such that cos y = x.
In other words, the inverse cosine function
cos^{-1 }: [-1, 1] --> [0, π]
is defined by cos^{-1}(x) = y if only only if cos y = x and y ∈ [0, y]
Principal domain :
The restricted domain [0, π] is called the principal domain of cosine function and the values of y = cos^{-1}(x), -1 ≤ x ≤ 1 are known as principal values of the function y = cos^{-1}(x).
From the definition of y = cos^{-1}(x), we observe the following :
i) y = cos^{-1}(x), if and only if x = cos y for -1 ≤ x ≤ 1 and 0 ≤ y ≤ π
ii) cos [cos^{-1}(x)] = x, if |x| ≤ 1 and has no sense if |x| > 1
iii) cos^{-1}(cos x) = x, if 0 ≤ x ≤ π, the range of cos^{-1} x.
Note that cos^{-1} (3π) = π
Find all values of x such that
Problem 1 :
-6π ≤ x ≤ 6π and cos x = 0
Solution :
cos x = 0
x = cos^{-1}(0)
For what angle measure of cosine, we get the value 0.
General solution for cosine function for cos^{-1}(0) :
Problem 2 :
-5π ≤ x ≤ 5π and cos x = 1
Solution :
cos x = 1
x = cos^{-1}(1)
For what angle measure of cosine, we get the value 1.
General solution for cosine function for cos^{-1}(1) :
x = 2nπ, where n ∈ N
If n = 0 x = 2(0)π x = 0 |
If n = 1 x = 2(1)π x = 2π |
If n = -1 x = 2(-1)π x = -2π |
If n = -2 x = 2(-2)π x = -4π |
If n = 2 x = 2(2)π x = 4π |
If n = -3 x = 2(-3)π x = -6π |
If n = 3
x = 2(3)π
x = 6π∉-5π ≤ x ≤ 5π
So, the possible values of n are
0, ±1, ±2, -3
Problem 3 :
State the reason for
Solution :
Here the angle measure (-π/6) mentioned is not in the domain [0, π]. So, it is not true.
Problem 4 :
Is cos^{-1}(-x) = π - cos^{-1}(x) true ? justify your answer ?
Solution :
cos^{-1}(-x) = π - cos^{-1}(x)
Here the value mention is negative. Using ASTC formula in 2n quadrant and in 3rd quadrant cosine will be negative. Considering the domain [0, π].
To get the angle, we should use
π - cos^{-1}(x)
So, the given statement is true.
Problem 5 :
Find the principal value of cos^{-1} (1/2)
Solution :
cos^{-1} (1/2)
For what angle measure of cosine, we get 1/2.
For 60 degree, we get 1/2 and it lies in the domain [0, π].
cos^{-1} (1/2) = π/3
Find the value of
Problem 6 :
2cos^{-1}(1/2) + sin^{-1}(1/2)
Solution :
= 2cos^{-1}(1/2) + sin^{-1}(1/2)
= 2 (π/3) + π/6
= 2π/3 + π/6
= (4π + π)/6
= 5π/6
Problem 7 :
cos^{-1}(1/2) + sin^{-1}(-1)
Solution :
For what angle measure of cosine, we get 1/2 and for what angle measure of sin, we get -1.
= (π/3) + (3π/2)
Here (π/3) is in the domain of cosine. But 3π/2 is not in the domain of sin.
So, -π/2 is the replacement of 3π/2.
= (π/3) + (-π/2)
= (2π - 3π)/6
= -π/6
Problem 8 :
Solution :
Problem 9 :
Solution :
Combining the above domain, we get [-5, 5]
So, the domain of the function f(x) is [-5, 5].
Problem 10 :
For what value of x, the inequality
π/2 < cos^{-1} (3x- 1) < π
holds ?
Solution :
To solve for x,
π/2 < cos^{-1} (3x- 1) < π
cos (π/2) < (3x - 1) < cos π
0 < 3x - 1 < -1
Add 1
1 < 3x < 0
Divide by 3
1/3 < x < 0/3
1/3 < x < 0
The inequality is true when,
0 < x < 1/3
Find the value of
Problem 10 :
Solution :
Problem 11 :
Solution :
May 21, 24 08:51 PM
May 21, 24 08:51 AM
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